No Slide Title · synonymous with molar mass and ... elements are present and the simplest whole-number ratio of their atoms. ... determine the smallest whole number ratio by

Molecular Mass

synonymous with molar mass and molecular weight

Molecular Mass

is the sum of the atomic masses of all the atoms in a molecule

the mass in grams of one mole of a compound

not all compounds are molecular

Formula Mass

calculated exactly the same way as molecular mass

Solid structure of NaCl

Determine the molar mass of Sr(NO3)2 .

pg. 322 problem 25

1 mol Sr(NO3)2

2 mol N = 2(14.00 g)1 mol Sr = 87.62g

6 mol O = 6(16.00 g)

= 211.62 g

pg. 323 problem 27What is the mass in grams of 3.25 mol sulfuric acid ( H2SO4 ).

3.25 mol H2SO4

98.07 g H2SO4

1 mol H2SO4

x = 319 g H2SO4

1 mol H2SO4

2 mol H = 2(1.00 g)1 mol S = 32.07g

4 mol O = 4(16.00 g)

= 98.07 g

pg. 324 problem 30aDetermine the number of moles present in 22.6 g AgNO3.

22.6 g AgNO3

1 mol AgNO3

169.87 g AgNO3

x = 0.133 mol AgNO3

1 mol AgNO3

1 mol N = 14.00 g1 mol Ag = 107.87g

3 mol O = 3(16.00 g)

= 169.87 g

pg. 326 problem 31aA sample of silver chromate (Ag2CrO4 ) has a mass of 25.8 g.

25.8 g Ag2CrO4

1 mol Ag2CrO4

331.74 g Ag2CrO4 x

= 9.36 x 1022 ions Ag+

1 mol Ag2CrO4

1 mol Cr = 52.00 g

2 mol Ag = 107.87g

4 mol O = 4(16.00 g)

= 331.74 g

a. How many Ag+ ions are present?

2 mol Ag+

1 mol Ag2CrO4

x

6.02 x 1023 ions Ag+

1 mol Ag+

x

pg. 326 problem 31bA sample of silver chromate (Ag2CrO4 ) has a mass of 25.8 g.

25.8 g Ag2CrO4

1 mol Ag2CrO4

331.74 g Ag2CrO4 x

= 4.68 x 1022 atoms CrO42-

1 mol Ag2CrO4

1 mol Cr = 52.00 g

2 mol Ag = 107.87g

4 mol O = 4(16.00 g)

= 331.74 g

b. How many CrO42- ions are present?

1 mol CrO42-

1 mol Ag2CrO4

x

6.02 x 1023 ions CrO42-

1 mol CrO42-

x

pg. 326 problem 31cc. What is the mass in grams of one formula unit of silver chromate (Ag2CrO4 ).

1 mol Ag2CrO4

1 mol Cr = 52.00 g

2 mol Ag = 107.87g

4 mol O = 4(16.00 g)

= 331.74 g

1 mol Ag2CrO4

331.74 g Ag2CrO4 1 mol Ag2CrO4x

6.02 x 1023 f.u. Ag2CrO4

g Ag2CrO4

f.u. Ag2CrO4

=

5.51 x 10-22

Percent Composition of Compounds

Percent composition is the percent by mass of each element the compound contains.

Obtained by dividing the mass of each element in one mole of the compound by the molar mass of the compound and multiplying by 100%

Example 45

Calculate the percent composition by mass of H,P and O for one mole of phosphoric acid (H3PO4)

Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99

3.086%

Molar mass = 3(1.008g) + 30.97g + 4(16.00) = 97.99

31.61%

65.31%

x 100% =%H = 97.99g

3(1.008g)

100% =%P = x30.97g

97.99g

100% =%O = x4(16.00)97.99g

ExampleA sample of a compound containing carbon and oxygen had a mass of 88g. Of this sample 24g was carbon, 64g was oxygen. What is the percent composition of this compound.

100% =%oxygen = 88g

x64g 73%

100% =%carbon = 88g

x24g27%

Determining Formula

Elemental Composition

Levels of Structure

Empirical Formula

Molecular FormulaConstitution

ConfigurationConformation

Examples:


Formaldehyde Glucose

C: 40.00% C: 40.00%H: 6.73% H: 6.73%O: 53.27% O: 53.27%


Empirical Formula



Levels of Structure

√

The empirical formula tells us which elements are present and the simplest whole-number ratio of their atoms.

Empirical Formula

Examples: Formaldehyde and Glucose

Empirical Formula

Elemental CompositionC: 40.00%H: 6.73%O: 53.27%

assume a 100g samplecalculate atom ratios by dividing by atomic weight

40.00 g6.73 g

53.27 g

40.00 g x 1 mol12.01 g

= 3.33 mol

6.73 g x 1 mol1.00 g

= 6.73 mol

53.27 g x 1 mol16.0 g

= 3.33 mol

100 g

Calculating Empirical Formula

C:

H:

O:

Examples: Formaldehyde and GlucoseEmpirical Formula


assume a 100g sample

calculate atom ratios by dividing by atomic weight

40.00 g6.73 g

53.27 g

3.33 mol6.73 mol3.33 mol

determine the smallest whole number ratio by dividing by the smallest molar value

40.00 g x 1 mol12.01 g

= 3.33 mol

6.73 g x 1 mol1.00 g

= 6.73 mol

53.27 g x 1 mol16.0 g

= 3.33 mol

100 g

Calculating Empirical Formula

3.33 mol

3.33 mol

3.33 mol

= 1.00

= 2.02

= 1.00

C:

H:

O:

Examples: Formaldehyde and Glucose

Empirical Formula


40.00 g6.73 g

53.27 g

3.33 mol6.73 mol3.33 mol

121

Empirical Formula: CH2O

A 1.723 g sample of aluminum oxide (which consists of aluminum and oxygen only) contains 0.912g of Al. Determine the empirical formula of the compound.

Example

0.912 g Al = 0.811 g O 1.723 g sample -

0.912 g Al x1 mol Al

26.98 g Al

0.811 g O x1 mol O

16.0 g O

= 0.0338 mol

= 0.0507 mol

0.0338 mol

0.0338 mol

= 1.0

1.5=

x 2 =

x 2 =

Al2O3

2

3


Empirical Formula



Levels of Structure

√√

determined from empirical formula and experimentally determined molecular mass

Molecular Formula

Compound EmpiricalFormula

Molar mass

formaldehyde CH2Oglucose CH2O

30180

Calculation of empirical mass1 mol C = 12.01 g2 mol H = 2 x 1.016 g1 mol O = 16.00g

30.026g

30g 30g

= 1formaldehyde

180g 30g

= 6glucose

Molecular mass

Empirical mass

determined from empirical formula and experimentally determined molecular mass

Molecular Formula


Molar mass

Molecularformula

formaldehyde CH2Oglucose CH2O

30180

CH2OC6H12O6

Example:


LysineC: 49.20%H: 9.66%

O: 21.94%N: 19.20%


Empirical Formula



√

Levels of Structure

Example:

Elemental Formula

LysineC: 49.20%H: 9.66%

O: 21.94%N: 19.20%

assume a 100-g sample

49.20 g9.66 g19.20 g21.94 g

calculate atom ratios by dividing by atomic weight

determine smallest whole-number ratioby dividing by smallest number

49.20 g C

9.66 g H

19.20 g N

21.94 g O x1 mol

16.0g= 1.37 mol1.37 mol

x1 mol12g

4.10 mol=

9.58 molx1 mol1.0g

=

1.37 molx1 mol

14.0g=

1.37 mol

1.37 mol

1.37 mol

1.37 mol=

=

=

= l

l

7

3

Example:

Elemental Formula (cont’d)

LysineC: 4.10 mol C atomsH: 9.58 mol H atoms

O: 1.37 mol O atomsN: 1.37 mol N atoms

determine smallest whole-number ratioby dividing by smallest number (1.37 mol)

3711

C3H7ON


Empirical Formula



√√

Levels of Structure

determined from empirical formula and molar mass

Molecular Formula


Molar mass

Molecularformula

lysine C3H7ON ~150 C6H14O2N2

73g

150g= 2

Write the empirical formulas for the following molecules: (a) acetylene (C2H2), (b) dinitrogen tetroxide (N2O4), (c) glucose (C6H12O6), diiodine pentoxide (I2O5).

Example

This problem is not realistic. Molecular formulas are derived from empirical formulas, not vice versa. Empirical formulas come from experiment.

Calculate the number of moles of chloroform (CHCl3) in 198 g of chloroform.

Example

Molecular mass of chloroform:1 mol C = 12.01 g1 mol H = 1.008 g3 mol Cl = 3(35.46 g) = 106.38 g

1 mol CHCl3 = 119.4 g

= 1.66 mol CHCl3198 g CHCl3 x1 mol CHCl3

119.4 g CHCl3

Caffeine contains 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of 194.2g. Determine the molecular formula formula of caffeine

Example

First determine the mass of each element in one mole of caffeine

100g caffeine

49.48g C

5.15g H

100g caffeine

28.87g N

100g caffeine

16.49g O

100g caffeine

Example

x194.2g caffeine

1 mol

x 194.2g caffeine

1 mol

x 194.2g caffeine

1 mol

x 194.2g caffeine

1 mol

=1 mol caffeine

96.09g C

=1 mol caffeine

10.0g H

=1 mol caffeine

56.07g N

=1 mol caffeine

32.02g O

then convert to moles

Example

1 mol caffeine

96.09g C

1 mol caffeine

10.0g H

1 mol caffeine

56.07g N

1 mol caffeine

32.02g O

x

x

x

x

12.011g C

1 mol C

1.008g H

1 mol H

14.01g N

1 mol N

16.00g O

1 mol O

1 mol caffeine

8.00 mol C=

1 mol caffeine

4.00 mol N=

1 mol caffeine

2.00 mol O=

1 mol caffeine

9.92 mol H=

C8H10N4O2

Documents

No Slide Title · synonymous with molar mass and ... elements are present and the simplest whole-number ratio of their atoms. ... determine the smallest whole number ratio by